$A$ manufacturer has $600 \text{ litres}$ of a $12 \%$ solution of acid. How many litres of a $30 \%$ acid solution must be added to it so that the acid content in the resulting mixture will be more than $15 \%$ but less than $18 \%$?

Let $x$ be the number of litres of the $30 \%$ acid solution to be added.
The total volume of the mixture will be $(600 + x) \text{ litres}$.
The amount of acid in the mixture is $(0.12 \times 600 + 0.30 \times x) = (72 + 0.3x) \text{ litres}$.
According to the problem,the acid content must be more than $15 \%$ and less than $18 \%$ of the total volume:
$0.15(600 + x) < 72 + 0.3x < 0.18(600 + x)$.
Solving the first inequality: $90 + 0.15x < 72 + 0.3x \implies 18 < 0.15x \implies x > 120$.
Solving the second inequality: $72 + 0.3x < 108 + 0.18x \implies 0.12x < 36 \implies x < 300$.
Thus,the amount of $30 \%$ acid solution to be added must be more than $120 \text{ litres}$ and less than $300 \text{ litres}$.